16.

20 - Structural Mechanics
Spring 2012
Stress and equilibrium
Concept Questions #2 - Corrections
Instructor: Raúl Radovitzky
Aeronautics & Astronautics
M.I.T

Problem 1.1 (Stress states on two sets of faces). The plane stress state at a point is known
and characterized by the following stress tensor:
   
σ11 σ12 250 0
σ= =
σ21 σ22 0 250

in a coordinate system E = (e1 , e2 ), as illustrated in Figure 1.

∗ ∗ ∗
1. Determine the stress components σ11 , σ22 , σ12 in a coordinate system E ∗ = (e∗1 , e∗2 ),
where e∗1 is oriented at an angle of 25 degree with respect to e1 .

Solution: The stress tensor σ is given by

   
σ11 σ12 250 0
σ= = .
σ21 σ22 0 250

Herein σ11 = σ22 and σ12 = 0, hence this tensor corresponds to an hydrostatic state
of stress meaning that the stresses acting on a face with any arbitrary orientation
∗ ∗ ∗ ∗
are given by σ12 = σ21 = σ12 = σ21 = 0, and σ11 = σ22 = σ11 = σ22 = p, where p
is the so-called hydrostatic pressure. NB: the result is the same for any value of the
angle θ.
This problem can be also solved with the approach performed in the Concept Ques-
tion 1.2.2 on “Stresses on an inclined face”. We know that the stress vectors acting
on the faces perpendicular to the axes e1 , and e2 are defined by

t1 = σ11 e1 + σ12 e2 ,
t2 = σ21 e1 + σ22 e2 ,

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where σij are the component of the stress tensor.

After writing the axes e∗1 and e∗2 in the componentes of E = (e1 , e2 ), i.e., e∗1 =
cos θ e1 + sin θ e2 , and e∗2 = − sin θ e1 + cos θ e2 , we can calculate the tractions in
the directions e∗1 and e∗2 as

te∗1 = t1 cos θ + t2 sin θ = (σ11 e1 + σ12 e2 ) cos θ + (σ21 e1 + σ22 e2 ) sin θ,

te∗2 = −t1 sin θ + t2 cos θ = − (σ11 e1 + σ12 e2 ) sin θ + (σ21 e1 + σ22 e2 ) cos θ.

These expressions and the fact that σ11 = σ22 = p and σ12 = σ21 = 0 enable us to
compute
∗
σ11 = te∗1 · e∗1 = σ11 cos2 θ + σ22 sin2 θ + 2σ12 sin θ cos θ = p,
∗
σ22 = te∗2 · e∗2 = σ11 sin2 θ + σ22 cos2 θ − 2σ12 sin θ cos θ = p,
∗
= te∗1 · e∗2 = (σ22 − σ11 ) sin θ cos θ + σ12 cos2 θ − sin2 θ = 0.


σ12

We can achieve the same conclusion through a rotation of stresses (see the statement
of the Problem 1.2).

e∗2 e2

e∗1

θ
e1

Figure 1: Coordinate systems E = (e1 , e2 ) and E ∗ = (e∗1 , e∗2 ), where θ = 25◦ .

Problem 1.2 (Stress rotation formulae in matrix form). Specialize the general expression for
the transformation of stress components Equation (1.11) in the class notes to two dimensions
and show that they can be expressed in the following two ways:
 ∗    
 σ11  cos2 θ sin2 θ 2 sin θ cos θ  σ11 
σ∗ = sin2 θ cos2 θ −2 sin θ cos θ  σ22
 22∗
σ12 − sin θ cos θ sin θ cos θ cos2 θ − sin2 θ σ12
  

1. Show that this formula can be recast in the following compact matrix form
 ∗ ∗
    
σ11 σ12 cos θ sin θ σ11 σ12 cos θ − sin θ
∗ ∗ =
σ12 σ22 − sin θ cos θ σ12 σ22 sin θ cos θ

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Solution: In the solution of the Problem 1.1 we explain how to derive the rela-
∗ ∗ ∗
tion between the stress components σ11 , σ22 , σ12 expressed in a coordinate system
∗ ∗ ∗
E = (e1 , e2 ), and the stress components σ11 , σ22 , σ12 expressed in a coordinate sys-
∗ ∗ ∗ T
tem E = (e1 , e2 ). That relation is written here in matrix form: {σ11 σ22 σ12 } =
T
A {σ11 σ22 σ12 } .
Although the matrix A seems to represent a rotation transformation between two
vectors, it actually represents the
 ∗ rotation
 transformation between two tensors (re-
∗
σ σ12
member that the tensor σ ∗ = 11 ∗ ∗ is symmetric, and thus it can be written
σ12 σ22
as a vector turning to the Voigt notation). It means that the relation between
∗ ∗ ∗
{σ11 σ22 σ12 } and {σ11 σ22 σ12 } can be also expressed as a change of coordinates
between the tensors σ ∗ and σ, i.e.

σ ∗ = RT σR,
 
cos θ − sin θ
where R = in this case.
sin θ cos θ
It is clear now why the plane stress rotation formula and the compact matrix form
are equivalent.

Problem 1.3 (Principal stresses). Let’s consider the following state of stress:
   
σ11 σ12 σ13 200 50 −80
σ =  σ21 σ22 σ23  =  50 300 100  .
σ31 σ32 σ33 −80 100 −100
1. Determine both, the principal stresses and the principal directions.

Solution: We have a stress tensor σ given by

 
200 50 −80
σ =  50 300 100  .
−80 100 −100

To determine (1) the principal stresses and (2) the principal directions we have to
compute the eigenvalues and eigenvectors, respectively. After the calculation, we
obtain a stress tensor σ ∗ with the eigenvalues in the diagonal, and a matrix M ∗ with
the corresponding eigenvectors in columns:
 
331.64 0 0
σ∗ =  0 215.12 0 ,
0 0 −146.76

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and  
−0.2562 −0.9335 0.2508
M ∗ =  −0.9510 0.1970 −0.2381  .
−0.1729 0.2995 0.9383

We can calculate the invariants I1 , I2 and I3 of the stress tensor to verify is these
results are correct. We obtain I1 = I1∗ = 400, I2 = I2∗ = −8900, and I3 = I3∗ =
−10470000. It means that is the same tensor but expressed in different coordinate
systems. As one of the matrices is diagonal, we can conclude that is the matrix of
the principal stresses.

Problem 1.4 (Principal stresses and transformation). Let’s consider the following state of
stress:  
3 1 1
σ =  1 0 2 .
1 2 0
1. Using the stress invariants, determine both, the principal stresses and directions.
p
2. Determine the traction vector on a plane with a unit normal n = (0, 1, 1)/ (2).

Solution: 1. First, we calculate the three invariants:

I1 = tr(σ)
= σkk
= σ11 + σ22 + σ33
= 3
I2 = tr(σ −1 )det(σ)
1
= (σii σjj − σij σji )
2
2 2 2
= σ11 σ22 + σ22 σ33 + σ11 σ33 − σ12 − σ23 − σ13
= −6
I3 = det(σ)
2 2 2
= σ11 σ22 σ33 + 2σ12 σ23 σ31 − σ12 σ33 − σ23 σ11 + σ13 σ22
= −8

This leads to the following characteristic equation:

det[σij − λδij ] = −λ3 + I1 λ2 − I2 λ + I3 = 0

= −λ3 + 3λ2 + 6λ − 8 = 0

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The roots of this equation are found to be λ = 4, 1 and −2. Back-substituting the
first root into the fundamental system gives:
(1) (1) (1)
−n1 + n2 + n3 = 0
(1) (1) (1)
n1 − 4n2 + 2n3 = 0
(1) (1) (1)
n1 + 2n2 − 4n3 = 0

Solving this system,

p the normalized principal direction is found to be
n(1) = (2, 1,
p 1)/ (6). In similar fashion,
p the other two principal directions are n(2) =
(−1, 1, 1)/ (3) and n(1) = (0, −1, 1)/ (2).
2. The traction vector on the specified plane is calculated by using the relation:
    √ 
3 1 1 0√ 2/√2
Tin =  1 0 2   1/√2  =  2/√2  .
1 2 0 1/ 2 2/ 2

Problem 1.5 (Stress invariants for plane stress state). Let’s introduce the following two
quantities:

I1 = σ11 + σ22 ,
2
I2 = σ11 σ22 − σ12 .

1. In the case of plane stress problems, show that these two quantities are invariant.

2. Prove this invariance by showing that these quantities are identical when computed in
terms of the principal stresses and in terms of stresses acting on a face at an arbitrary
orientation.

Solution: The invariants of the stress tensor are the quantities that remain un-
changeable under coordinate transformations. In the three-dimensional case they
are defined as

I1 = σ11 + σ22 + σ33 ,

2 2 2
I2 = σ11 σ22 + σ22 σ33 + σ11 σ33 − σ12 − σ23 − σ13 ,
2 2 2
I3 = σ11 σ22 σ33 − σ11 σ23 − σ22 σ13 − σ33 σ12 + 2σ12 σ13 σ23 .

In the plane stress approach we assume that σ33 = 0, σ13 = 0, and σ23 = 0. By
replacing these stresses in the above equations we obtain the invariants for the plane

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16.20, Spring 2012 Concept Questions #2 - Corrections

stress state

I1 = σ11 + σ22 ,
2
I2 = σ11 σ22 − σ12 ,
I3 = 0,

which correspond with the values indicated in the statement of the problem.
Given the stress components σ11 , σ22 and σ12 on two orthogonal faces, we can write
p p
the principal stresses σ11 and σ22 as

p σ11 + σ22
σ11 = + ∆,
2
p σ11 + σ22
σ22 = − ∆,
2
q
σ11 −σ22 2

2
where ∆ = 2
+ σ12 .
These expressions make straightforward the following calculations

I1p = σ11
p p
+ σ22 = σ11 + σ22 ⇒ I1p = I1 .
p 2
I2p = σ11
p p
σ22 − (σ12 ) = σ11 σ22 − (σ12 )2 ⇒ I2p = I2 .

If the stress components σ11 , σ22 and σ12 on two orthogonal faces are known, the
stresses acting on a face with an arbitrary direction θ can be calculated as
 
∗ σ11 + σ22 σ11 − σ22
σ11 = + cos 2θ + σ12 sin 2θ,
2 2
 
∗ σ11 + σ22 σ11 − σ22
σ22 = − cos 2θ − σ12 sin 2θ,
2 2
 
∗ σ11 − σ22
σ12 = − sin 2θ + σ12 cos 2θ.
2

These expressions make straightforward the following calculations

I1∗ = σ11
∗ ∗
+ σ22 = σ11 + σ22 ⇒ I1∗ = I1 .
∗ 2
∗ ∗
I2∗ = σ11 σ22 − (σ12 ) = σ11 σ22 − (σ12 )2 ⇒ I2∗ = I2 .

Problem 1.6 (Mohr’s circle derivation). Let’s consider a Mohr’s circle of radius R and
∗ ∗
principal stresses σ11 and σ22 which are points belonging to both, the circle and the σii axis.
1. Derive the expression of the radius R as a function of the stress components.
2. Derive the expressions of the principal stresses.

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Solution: 1. In Figure 2, from the Pythagore theorem we have:

2 2
σm + σ12 = R2

and
σ11 − σ22
σm =
2
hence:  2
σ11 − σ22 2
+ σ12 = R2
2
finally we obtain the following relation for the radius of the circle:
s 2
1 2
R= (σ11 − σ22 ) + σ12
2

2. The value of principal stresses is equal to the ordinate of the origin ( 21 (σ11 + σ22 ))
of the Mohr’s circle + or − the radius of the circle R:
s 2
∗ 1 1 2
σ11 = (σ11 + σ22 ) + (σ11 − σ22 ) + σ12
2 2
s 2
∗ 1 1 2
σ22 = (σ11 + σ22 ) − (σ11 − σ22 ) + σ12 (1)
2 2

σij

(σ11 ,σ12 ) b

σ22
∗ σm σ11
∗

b b b σi
2θ ∗

R
b

(σ22 ,−σ12 )

Figure 2: Mohr circle.

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Problem 1.7 (Mohr’s circle). Let’s consider the following state of stress:
 
80 40
σ= .
40 −20
1. Draw the Mohr’s circle of this state of stress
2. Using the Mohr’s circle, determine the principal stresses and the corresponding direc-
tions.
3. Using the Mohr’s circle, calculate the stresses on axes rotated 60 degrees counterclockwise
from the reference axes.
4. Compare these results with the ones obtained analytically?

Solution: The stress tensor σ is given by

   
σ11 σ12 80 40
σ= = .
σ21 σ22 40 −20

1. We obtain the Mohr’s circle (Figure 3) by plotting the two points with the co-
ordinates (σ22 ,−σ12 ) and (σ11 ,σ12 ), respectively. The line between these two points
intersects the σii −axis at the center of the circle.

2. The principal stresses are the stresses for which the shear components are zero,
∗ ∗
i.e when the stress matrix is diagonal. Hence, σ11 and σ22 are the two intersection
∗
points between the circle and the σii −axis. These stresses are (σ11 )plot ≈ 94 MPa and
∗
(σ22 )plot ≈ −34 MPa, respectively. The direction is given by the angle 2θ between
the line joining (σ22 ,−σ12 ) and (σ11 ,σ12 ) and the σii −axis. Herein, (θ∗ )plot ≈ 19o .

3. For an angle 2θ equals to 60o from the reference axis (or σii −axis), we obtain the
stresses σ11 ' 39.6 MPa, σ22 ' 20.3 MPa and σ12 ' −63.3 MPa.

4. The principal stresses and corresponding directions are calculated analytically

with the following equations

∗ σ11 + σ22
σ11 = + ∆,
2
∗ σ11 + σ22
σ22 = − ∆,
2
2σ12
tan 2θp =
σ11 − σ22
q
σ11 −σ22 2

2
where ∆ = 2
+ σ12 is the shear stress component, around 0 when the
stress tensor is expressed in the principal directions system of coordinates.

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s 2
80 − −20
∆ = + (40)2
2
= 64.0312 MPa

so
∗ 80 − 20
σ11 = + 64.0312
2
= 94.0312 MPa

and
∗ 80 − 20
σ22 = − 64.0312
2
= −34.0312 MPa

and
2 × 40
tan 2θp =
80 + 20
= 0.8
p p
Hence, we obtain σ11 = 94.0312, σ22 = −34.0312, and θp = 19.33, and these results
are in agreement with our previous results.

σij
(σ11 , σ12 )
b

b
(σ11 ,σ12 )

σ22
∗
= −34 MPa 2θ σ11
∗
= 94 MPa
2θ∗
b b b σi

(σ22 ,−σ12 ) b

(σ22 , −σ12 )

Figure 3: Mohr circle.

Problem 1.8 (Stress and equilibrium). Let’s consider an elastic structural member for which

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the stress field is expressed as follows:

 
−x31 + x22 5x3 + 2x22 x1 x33 + x21 x2
σ =  5x3 + 2x22 2x31 + 21 x22 0 .
3 2 2 3
x1 x3 + x1 x2 0 4x2 − x3
1. Determine the body force distribution for equilibrium in static.

Solution: We apply the following relation of equilibrium:

∇·σ+F =0

∂σ11 ∂σ12 ∂σ13

 
+ +
 ∂x1 ∂x2 ∂x3 
 ∂σ21 ∂σ22 ∂σ23 
∇·σ = + +
 
∂x1 ∂x2 ∂x3

 
 ∂σ31 ∂σ32 ∂σ33 
+ +
∂x1 ∂x2 ∂x3
Which leads to:

−3x21 + 4x2 + 3x1 x23 + F1 = 0

x2 + F 2 = 0
−3x23 + x33 + 2x1 x2 + F3 = 0

The body force distribution, as obtained from these expressions, is therefore:

F1 = 3x21 − 4x2 − 3x1 x23

F2 = −x2
F3 = −2x1 x2 − x33 + −3x23

The state of stress and body force at any specific point within the member may
be obtained by substituting the specific values of x1 , x2 and x3 into the previous
equations.

Problem 1.9 (Stress fields in static equilibrium). Let’s consider a structure in equilibrium
and free of body forces. Are the following stress fields possible?
x22
 
 c1 x1 + c2 x2 + c3 x1 x2 −c3 2 − c1 x2 
1. σ =  .
x22
−c3 − c1 x2 c4 x 1 + c1 x 2
2
 
3x1 + 5x2 4x1 − 3x2
2. σ = .
4x1 − 3x2 2x1 − 4x2

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 
x21 − 2x1 x2 + cx3 −x1 x2 + x22 −x1 x3
3. σ =  −x1 x2 + x22 x22 −x2 x3 .
−x1 x3 −x2 x3 (x1 + x2 )x3

Solution: To solve this problem we turn to the momentum equation

∂σji ∂ 2 ui
+ ρfi = ρ 2 .
∂xj ∂t

As the structure is in equilibrium (steady state) and free of body forces, the terms
2
ρ ∂∂tu2i and ρfi are null. Then, the equilibrium equations become

∂σji
= 0.
∂xj

For a 2D stress field we have:

∂σ11 ∂σ12
 
+
∂σji  ∂x 1 ∂x2 
= (2)
 
∂xj

 ∂σ ∂σ22 
21
+
∂x1 ∂x2

For a 3D stress field we have:

∂σ11 ∂σ12 ∂σ13
 
 ∂x1 + ∂x2 + ∂x3 
 
 
∂σji 
 ∂σ21 ∂σ22 ∂σ23

= + + (3)

∂xj  ∂x1 ∂x2 ∂x3


 
 
 ∂σ31 ∂σ32 ∂σ33 
+ +
∂x1 ∂x2 ∂x3

1. For the first stress field we obtained:

   
∂σji c1 + c3 x2 + −c3 x2 − c1 0
= = (4)
∂xj c1 c1
This stress field does not satisfy the equilibrium equation.
2. For the second stress field we obtained:
   
∂σji 3−3 0
= = (5)
∂xj 4−4 0
This stress field does satisfy the equilibrium equation.

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3. For the third stress field we obtained:

   
2x1 − 2x2 − x1 + 2x2 − x1 0
∂σji 
= −x2 + 2x2 − x2 = 0  (6)
∂xj
−x3 − x3 + x1 + x2 −2x3 + x1 + x2
This stress field does not satisfy the equilibrium equation

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